The bond enthalpy of N2(g) is 418.0 kJ/mol. Calculate delta H°f for N(g).?
To calculate ΔH°f for N(g), we need to use Hess's law and the known bond enthalpy of N2.
Hess's law states that the enthalpy change of a reaction is the same, regardless of the pathway taken. In this case, we can use the following chemical equation to calculate ΔH°f for N(g):
1/2 N2(g) → N(g)
Since we have the bond enthalpy of N2 (418.0 kJ/mol), we need to consider that breaking one mole of N2 bonds requires an input of energy (endothermic), while forming half a mole of N-N bonds releases energy (exothermic).
The enthalpy change for the bond breaking (endothermic) is equal to the bond enthalpy:
ΔH° = +418.0 kJ/mol
The enthalpy change for bond formation (exothermic) is equal to the negative of the bond enthalpy. Since we are forming half a mole, we multiply the bond enthalpy by 2:
ΔH° = -2 × (418.0 kJ/mol)
Now, we can calculate the ΔH°f for N(g) using Hess's law:
ΔH°f(N(g)) = Σ(ΔH°f(products)) - Σ(ΔH°f(reactants))
Since N(g) is the product, and there are no other reactants, the equation simplifies to:
ΔH°f(N(g)) = -2 × (418.0 kJ/mol)
Calculating this expression gives us:
ΔH°f(N(g)) = -836.0 kJ/mol
Therefore, the standard enthalpy of formation for N(g) is -836.0 kJ/mol.